For i = 1 , 2 , 3 and j = 1 , 2 , 3 . If a i 2 + b i 2 + c i 2 = 1 ,   a i a j + b i b j + c i c j = 0 …

For i=1,2,3 and j=1,2,3. If ai2+bi2+ci2=1, aiaj+bibj+cicj=0,ij and A=a1a2a3b1b2b3c1c2c3 then detAAT=
  1. 0
  2. 1
  3. -1
  4. 3

Solution

We know AAT=AAT=A2

i.e. a1a2a3b1b2b3c1c2c32=a1b1c1a2b2c2a3b3c3a1b1c1a2b2c2a3b3c3

=Σa12Σa1a2Σa1a3Σa2a1Σa22Σa2a3Σa3a1Σa3a2Σa32=100010001

=1

Asked in: AP EAMCET 2022 (04 Jul Shift 1)

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